Question

A block of mass $m$ is placed at the bottom of a massless smooth wedge which is placed on a horizontal surface. When we push the wedge with a constant force, the block moves up the wedge. The work done by the external agent when the block has a speed $v$ and has reached the top of the wedge is

• A. $\frac{1}{2}m{v}^2+mgh$
• B. $\frac{1}{2}m{v}^2-mgh$
• C. $-\frac{1}{2}m{v}^2+mgh$
• D. $-\frac{1}{2}m{v}^2-mgh$

Answer 1

The correct option is A $\frac{1}{2}m{v}^2+mgh$


Applying Work - Energy theorem for wedge-block system,
$W_{\text{ext}}+W_{\text{gravity}}=\Delta KE$
$⟹W_{\text{ext}}-mgh=\Delta KE$
This gives $W_{\text{ext}}=mgh+\Delta KE\dots \left(i\right)$

$\Delta KE=$ Change in $KE$ of the block
$=\frac{1}{2}m{v}^2\dots \left(ii\right)$
Using Eqs. (i) and (ii), we have $W_{\text{ext}}=\frac{1}{2}m{v}^2+mgh$